Calculations regarding transmitter flow rate, differential pressure, and output current
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I have a question: if a transmitter includes a square root function, what is the relationship between the output current and the differential pressure? For example: Use a 1151dp4e capacitive differential pressure transmitter to measure flow rate. The differential pressure Δp is 25 kPa (the maximum differential pressure), and the range of the secondary meter is 0–200 t/h. Determine the corresponding differential pressure and current values when the flow rate is 80 t/h. The answer in the book states that the pressure difference is 4 kPa, which is easy to understand; the flow rate is proportional to the square root of the pressure difference ； The current is 6.56 mA. How should this be understood? Is the differential voltage proportional to the current? If that’s the case, wouldn’t the differential pressure and flow rate also be in a proportional relationship? This area is a bit messy. What are the differences between the old meters and the smart meters nowadays?
Is that the set of questions for instrument technicians that you’re looking at? I remember that there were many mistakes in those calculations
The output current of the differential pressure transmitter corresponding to 4 kPa is 6.56 mA, that is: 4 (differential pressure) * 16 (transmitter current output range of 20 mA–4 mA) / 25 (range of differential pressure for the transmitter) + 4 (lower limit of the transmitter’s current output)
That’s correct; the flow rate is proportional to the square root of the pressure difference, as well as to the square root of (current value – 4). Instantaneous flow rate / full flow rate = square root of instantaneous pressure difference / square root of full-scale pressure difference = square root of current value / square root of 16. In this case, it’s 80/200 = square root of 4 / square root of 25 = square root of 2.56 / square root of 16
16/25x4+4=6.25
After taking the square root of the meter reading, simple arithmetic is performed: 1. Output current I = (80÷200)×16+4 = 10.4mA; 2. Output current I = (√4 ÷ √25)×16+4 = 10.4mA. It’s that simple to calculate
The last edit to this post was made by cqdfwy on 2019-6-5 at 11:39. The entire process from measuring the differential pressure with a transmitter to having the flow rate displayed on a secondary meter is as follows: the transmitter measures the differential pressure ΔP, which is then converted into a current value I after taking the square root; this current value I is sent to the secondary meter, where it is used to display the flow rate Q. In other words, there is a linear relationship between I and Q. It is known that the measurement range of the transmitter is ΔPmax = 25 kPa, and the display range of the secondary meter is Qmax = 200 t/h. It is necessary to calculate the differential pressure value ΔP measured by the transmitter and the output current value I of the transmitter when the secondary meter shows Q = 80 t/h. I is in a linear relationship with Q; therefore, (I–I0)/(Imax–I0) = Q/Qmax, where I0 is the current value corresponding to the zero point. Thus, I = Q/Qmax * (Imax–I0) + I0. Substituting the numerical values gives: The output current of the transmitter, I, = 80/200 * (20–4) + 4 = 10.4 (mA). The differential pressure measured by the transmitter, ΔP, is related to its output current I through a square root relationship, that is, √ΔP/√ΔPmax = (I–I0)/(Imax–I0). Hence, ΔP = [(I–I0)/(Imax–I0) * (I–I0)]² * ΔPmax. Substituting the numerical values yields: The differential pressure measured by the transmitter, ΔP, = [(10.4–4)/(20–4)]² * 25 = 4 (kPa)
If you don’t understand, it’s fine. Why go to such lengths to understand it?